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done

two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6.

The chord parallel to these chords and midway between them is of

length \(\sqrt{a}\). Find the value of a.

Phythagora's

\(\begin{array}{|lrcll|} \hline (1) & x^2+7^2 &=& r^2 \\ (2) & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ (3) & 5^2 + \Big(3+(3-x)\Big)^2 &=& r^2 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)=(3): & r^2 = x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + (6-x)^2 \\ & x^2+7^2 &=& 5^2 + 6^2-12x+x^2 \\ & 7^2 &=& 5^2 + 6^2-12x \\ & 12x &=& 5^2 + 6^2 - 7^2 \\ & 12x &=& 12 \quad | \quad : 12 \\ & \mathbf{x}&=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{r^2=\ ?} \\ \hline (1): & r^2 &=& x^2+7^2 \quad | \quad \mathbf{x=1} \\ & r^2 &=& 1^2+7^2 \\ & \mathbf{r^2}&=& \mathbf{50} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{a=\ ?} \\ \hline (2): & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ & (3-x)^2 + \dfrac{a}{4} &=& r^2 \quad | \quad \mathbf{x=1},\ \mathbf{r^2=50} \\ & (3-1)^2 + \dfrac{a}{4} &=& 50 \\ & 2^2 + \dfrac{a}{4} &=& 50 \\ & 4 + \dfrac{a}{4} &=& 50 \quad | \quad -4 \\ & \dfrac{a}{4} &=& 46 \quad | \quad * 4 \\ & \mathbf{a}&=& \mathbf{184} \\ \hline \end{array}\)

2.

x + xy^2  =  250y

x - xy^2   = -240y       add these  and we get that

2x =  10y

x = 5y

So we have  that

5y  + (5y)y^2 = 250y

5y^3  = 245y

5y^3  - 245y  =  0

5y  (y^2  - 49)   = 0

5y ( y - 7) ( y + 7)   = 0

Setting each factor to  0 and solving for y produces

y =0   y =  7   and  y   =-7

So   using  x = 5y....the solutions are

(x,y )  =   (0, 0)    ( 35, 7)   and (-35, - 7)

1.

ab^4  = 384     ⇒   a =  384/b^4 ⇒  a^2  = 384^2  / b^8     (1)

a^2b^5  = 4608     (2)

Sub  (1)  into (2)  for a^2   and we have  that

[(384^2) / b^8]  b^5 =  4608    simplify

384^2 / b^3   = 4608    rearrange  as

b^3   = 384^2 / 4608

b^3  = 32

b = ∛32   = ∛[8*4]  = 2∛4

And  

a = 384/ [ 2∛4]^4  =  384 / [16 * 4^(4/3) =   24 /  [ 4 * 4^(1/3) =    6 / ∛4  =  6∛4^2 / 4  = (3/2)∛16  = 

(3/2)*2 * ∛2  = 3∛2

So

(a,b)   =   ( 3∛2 , 2∛4)

No you did not fix it!

Thanks, Melody  and Guest   !!!!

No, it does not sound dumb. I am quite sure that Chris appreciations your comments just as all answerers do.

If you become a member your appreciation will be more recognized

Members who make a good name for themselves definitely get preferential treatment.

Okay, I know that this might sound really dumb, but when I saw that you answered my question, I got very excited because whenever I need math help, I always see your answers and they're the most helpful. I just appreciate that a lot, so thank you. 

The  form we  seek is  

y = Acsc ( Bx)  + C

A =  the amplitude  =  [5 - - 9] / 2  =  14/ 2  =  7

C  =  the midline  =  [ 5-9]/2 = -4/2 =   -2

The period  =  2pi -  0  =  2pi

And

B  =  normal period  /  our period         =    pi  / [ 2pi]   =   1/2

So....our equation  is

y =  7csc ( (1/2)x)  - 2

Here's  the graph  :  https://www.desmos.com/calculator/qpwj6t9ook

P(at least  one defective laptop)  =

1 - P(neither laptop is  defective)  = 

1  - (20/23)(19/22)   =   

63/253 ≈  .249  ≈  24.9%

OK....glad to help   (even though I forgot most of the Physics I once learned.....LOL!!! )

Thank you, very helpful :)

Note  that 

120km            1000m           hr

______ x       _______  x  _______  =    100/3  m/s

   hr                  1km            3600 s

We  have  that

Final velocity=  Initial Velocity +  acceleration * time

So

0  =  (100/3)m/s  +  -1.8  m / s^2  *  time

- (100/3)m/s  =  -1.8m/s^2  * time

(100/3)m/s  

__________  =   time

1.8 m/s^2

(100/3) s                     100

_______  =  time  =   _____ s    ≈  18.52 s

    1.8                           5.4

B

C    7√5          A

We  have  a 30-60-90  right triangle

Angle  CBA   = 60°

Angle  BCA  = 90°.....the side opposite this has the length   2 ( 7√5) / √3  = 14√5 / √3  = 14√15 / 3

Angle  BAC  = 30°  ....the  side opposite this has the length  7√15/3

So....the perimeter  = 

7√5  +  (√15/3) (14 + 7)  =

7√5  + (√15/3)(21)  =

7√5  + 7√15  =

7 (√5 + √15)  units ≈   42.76  units

Write a recursive rule for each sequence then find a9 {15/2, 6, 9/2, 3, 3/2,..}

We can write

15/2 , 12/2 , 9./2 , 6/2 , 3/2

The  recursive  rule  is

a_n+1  =  a_n - 3/2    where   a₁  =   15/2 

a₆  =  0

a₇ =  -3/2

a₈  =  -6/2

a₉  =  -9/2

Let ( h,k) be  the center  of the circle

Since  the square of the radius  =  the square of the radius we can set up this equation

(h - 1)^2  + ( k - 1)^2  =  (h - 1)^2  + (k - 3)^2     simplify

(k -1)^2  = (k - 3)^2 

k^2  - 2k + 1  =  k^2  - 6k + 9

-2k + 1 =  -6k + 9

4k  =  8

k  = 2

Then we have that

(h -1)^2 + (k-1)^2  =  ( h - 9)^2  + (k - 2)^2

Subbing in for k we have

(h -1)^2  + (2 -1)^2  = (h -9)^2  + (2-2)^2    simplify

h^2 - 2h + 1  + 1  =  h^2 - 18h + 81

-2h + 2 =  -18h+ 81

16h  =  79

h = 79/16

So  the center is  (79/16 ,2)

And  the  radius can be  found as

(79/16 - 1)^2  + (2 -1)^2  =  r^2

(63/16)^2  + 1  = r^2

4225/256  =  r^2

So the equation is

( x - 79/16)^2 + (y - 2)^2  = 4225/256

write a recursive rule for each sequence then find a9 {15/2, 6, 9/2, 3, 3/2,..}

The probability is:  ₁₄C₄ · (0.44)⁴ · (0.56)¹⁰

                          +  ₁₄C₃ · (0.44)³ · (0.56)¹¹

                          +  ₁₄C₂ · (0.44)² · (0.56)¹²

                          +  ₁₄C₁ · (0.44)¹ · (0.56)¹³  

                          +  ₁₄C₀ · (0.44)⁰ · (0.56)¹⁴  

Find this sum ...

x is directly proportional to y   --->      y  =  k·x     --->      y₁  =  k·x₁     --->     y₂  =  k·x₂  

y₁  =  k·x₁     --->     k  =  y₁ / x₁     

y₂  =  k·x₂     --->     k  =  y₂ / x₂   

Combining:     y₁ / x₁  =  y₂ / x₂   

Multiplying both sides by (x₁ / y₂)     --->         (x₁ / y₂​) · (y₁ / x₁)  =  (x₁ / y₂​) · (y₂ / x₂)      

                                                          --->                           y₁ / y₂  =  x₁ / x₂   

Since  x₁ / x₂  =  4                            --->                           y₁ / y₂  =  4

Therefore                                          --->                          y₂ / y₁  =  1/4

We   have

odd  even  odd even odd even  odd  =  4 * 3 * 3 * 2 * 2 * 1 * 1 =  144

Thank you moderator for not printing my answer to this posting... NOT!  S***w you.

Note  that  20 revs / min  =   20 * (35/2) * 2pi  = 20 * 35 pi  =  700 pi  cm / min

So....in  50 s....it travels    700 pi * (5/6)   ≈  1832.6 cm ≈  1833 cm

A composite polynominal p(x) can be decomposed into two or more polynomials; that is, there are

polynomials u(x) and v(x) for which p(x)  =  u( v(x) ).

Since the radius of the circle is 3 cm, the circumference of the circle is 2·pi·3, which is 6 pi, or about 18.85 cm.

The part of the circle that includes an arc length of 2 cm is:  2 cm / 18.85 cm, or about 0.11 of the circle.

So, if you want to find the area of that sector, find the total area of the circle and multiply that answer by 0.11

Not readable, change it to LaTeX or something that makes the question easier to understand clearly.